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MATH111C Abstract Algebra [111c]

Mar 30, 2026 - Present

Lecture 1. Field extensions [l1]

Mar 30, 2026

Definition 1.1. field extension [1201]

Let $F$ be a field. A field containing $F$ is called a field extensions of $F$ or an extension of $F$.

Notation. some notations [1201A]

We use

$$ K / F \qquad \text{or} \qquad \begin{aligned} &K \\ &| \\ &F \end{aligned} $$

to mean that $K$ is a field extension of $F$.

When considering field extensions of $F$, the field $F$ is sometimes called the base field.

Example 1.2. examples from characteristics [1202]

Every field of characteristic $0$ is a field extension of $\mathbb{Q}$.
Every field of characteristic $p$ is a field extension of $\mathbb{F}_{p}$.

Definition 1.3. degree of field extensions [1203]

1.3.1. degree [1203A]

If $K$ is a field extension of $F$, then $K$ can be viewed as an $F$-vector space.

Exegesis. vector space [1203AA]

TODO

The degree of $K / F$ is $$ [K : F] := \dim_{F}K .$$

1.3.2. finiteness [1203B]

A field extension $K / F$ is called finite if $[K : F] < \infty$.

Example 1.4. complex and real numbers, rationals and polynomials [1204]

$\mathbb{C}$ is a field extension of $\mathbb{R}$ and $[\mathbb{C} : \mathbb{R}] = 2$.
$\mathbb{R}$ is a field extension of $\mathbb{Q}$ and $[\mathbb{R} : \mathbb{Q}] = \infty$.
Let $F$ be a field and $p(x)$ be an irreducible polynomial of degree $\geq 1$ in $F[x]$. Then $K = F[x] / (p(x))$ is a field extension of $F$ and $[K : F] = \deg p(x)$.
$F = \mathbb{Q}, p(x) = x^{2} - 2, K = \mathbb{Q}[x] / (x^{2} - 2)$. Then $K \cong \mathbb{Q}[\sqrt{ 2 }] = \{ a + b\sqrt{ 2 } \mid a, b \in \mathbb{Q} \}$.
$[\mathbb{F}_{p}(t), \mathbb{F}_{p}] = \infty$. $1, t, t^{2}, \dots, t^{n}, \dots$ are linearly independent over $\mathbb{F}_{p}$, so $\dim_{\mathbb{F}_{p}}\mathbb{F}_{p}(t) = \infty$.

Remark 1.5. [1205]

Theorem 1.6. [1206]

Let $F$ be a field and $p(x)$ be an irreducible polynomial in $F[x]$. Then

$K = F[x] / (p(x))$ is a field extension of $F$,
$\theta = \overline{x} \in K$ is a root of $p(x)$ in $K$,
$[K : F] = \deg p(x)$. Let $n = \deg p(x)$. Then $$1, \theta, \theta^{2}, \dots, \theta^{n-1}$$ is a basis of $K$ as an $F$-vector space. (Hence, $K = \{ a_{0} + a_{1} \theta + \cdots + a_{n-1}\theta^{n-1} \mid a_{0}, \dots, a_{n-1} \in F\}$.)

Proof. [1206A]

TODO

Theorem 1.7. [1207]

Let $F \subseteq K \subseteq L$ be fields. Then $[L:F]$ is finite if and only if $[L:K]$ and $[K:F]$ are both finite, and $$ [L:F] = [L : K][K : F]. $$

Proof. sketch [1207A]

$\implies$: Given basis of $L$ over $K$ and $K$ over $F$, denote by $\alpha_{i}, 1 \leq i \leq n$ and $\beta_{j}, 1 \leq j \leq m$ respectively, then we can prove $\alpha_{i}\beta_{j} \in L, \forall i, j$ are linearly independent over $F$. It follows that $[L : F] \geq mn$. So if $[L : F] < \infty$, it must follow that $m < \infty$ and $n < \infty$.

$\impliedby$: Assume now $m, n < \infty$. We can prove that $\alpha_{i}\beta_{j} \forall i, j$ actually spans $L$ over $F$ by showing that every element of $L$ can be written in the form $\sum F \alpha_{i} \beta_{j}$. Then $\alpha_{i} \beta_{j}$ is a basis of $L$ over $F$, hence $[L : F] = mn < \infty$.

When $[L : F] = \infty$, at most one of $[L:K]$ and $[K:F]$ can be finite. For example, $[\mathbb{C} : \mathbb{Q}] = \infty$ and $[\mathbb{R} : \mathbb{Q}] = \infty$ while $[\mathbb{C} : \mathbb{R}] = 2 < \infty$.

Lecture 2. The subfield generated by a subset and simple extensions [l2]

Apr 1, 2026

Definition 2.1. generated fields [1301]

Let $K / F$ be a field extension and $S$ be a subset of $K$. Let $$ F(S) = \text{the intersection of all subfields of } K \text{ which contains }F\text{ and }S. $$ One can check that $F(S)$ is a subfield of $K$ and is the smallest subfield of $K$ that contains both $F$ and $S$. We call $F(S)$ the field generated by $S$ over $F$.

Notation. some notations [1301A]

When $S = \{ \alpha_{1}, \dots, \alpha_{n} \}$, we write $F(S)$ as $F(\alpha_{1}, \dots, \alpha_{n})$.

Given $S \subseteq K$, taking intersections is a general way to generate a smaller field extension of $F$, which also gives the smallest one that contains both $S$ and $F$.

From this definition, it is hard to compute $F(S)$ because there might be infinite subfields containing $F$ and $S$. So we need a workaround to get $F(S)$.

We know that taking field of fractions gives us a field, if we consider each element of $F(S)$ as a fraction of elements, what do we know about these elements?

They must belong to an integral domain.
They must be composed of elements from $F$ and $S$.

Interestingly, $K$ is a field, so taking fractions of elements from $S$ simply brings another element of $K$, and it also belongs to a subfield of $K$ containing $S$. Actually, to construct a subfield of $K$ containing $S$, we first need to take multiples of any elements of $S$, then take inverses of those multiples, next, take multiples of those inverses and elements of $S$.

Remember that the needed subfields also contain $F$, given a subfield containing $S$, how do we enlarge it so it contains $F$, too? Well, simply multiplying every element by elements of $F$ and taking sums of those multiples is a good way. Recall that $\operatorname{Frac}(F) = F$, hence we have the following proposition:

Proposition 2.2. the field of fractions of $F[S]$ is $F(S)$ [1302]

Given $K / F$ and $S \subseteq K$, let $$ F[S] = \{ \text{finite sums of }a \cdot \alpha_{1} \alpha_{2} \dots \alpha_{n} \mid a \in F, \alpha_{1}, \dots, \alpha_{n} \in S, n \in \mathbb{Z}_{\geq 0}\}. $$ Then $F[S]$ is a subring of $K$ and $$ F(S) = \left\{ \left. \frac{a}{b} \right\vert a, b \in F[S], b \neq 0 \right\}. $$ ($F(S)$ is the fraction field of $F[S]$, and it is exactly the same object as in generated fields.)

Exegesis 2.2.1. why finiteness [1302A]

TODO

Taking multiples ensures $F[S]$ is closed under multiplication. Taking finite sums then ensures $F[S]$ is closed under addition. Moreover, we can take $a = 1$ and $n = 0$, then $F[S]$ contains $1$. It is easy to verify that $F[S]$ is actually an integral domain, thus we can take its field of fractions.

Proposition 2.3. identity on generated fields [1303]

Let $\varphi : F(S) \to F(S)$ be a field homomorphism such that $\varphi(\alpha) = \alpha$ for all $\alpha \in F \cup S$. Then $\varphi = id_{F(S)}$.

Exegesis 2.3.1. why extends to $F(S)$ [1303A]

By the construction of $F(S)$ from elements of $F \cup S$ in Proposition 1302, obviously $\varphi = id_{F(S)}$.

Definition 2.4. finitely generated, simple extensions and primitive elements [1304]

A field extension $K / F$ is called finitely generated if there exist $\alpha_{1}, \dots, \alpha_{n} \in K$ such that $K = F(\alpha_{1}, \dots, \alpha_{n})$.
A field extension $K / F$ is called simple if $K = F(\alpha)$ for some $\alpha \in K$. In this case, $\alpha$ is called a primitive element for the field extension $K / F$.

Example 2.5. $F(\theta), \mathbb{F}_p(t) / \mathbb{F}_p, F(x)$ [1305]

Let $p(x)$ be an irreducible element in $F[x]$. Then by Theorem 1206, we know that $K = F[x] / \left( p(x) \right)$ is a field extension of $F$, and $K = F + F\theta + \cdots + F\theta^{n-1}$ with $\theta = \overline{x} \in F[x] / \left( p(x) \right)$. Hence $K = F[\theta] = F(\theta)$, and $K / F$ is a simple extension, and $\theta$ is a primitive element for $K / F$.
$\mathbb{F}_{p}(t) / \mathbb{F}_{p}$ is simple and $t$ is a primitive element for $\mathbb{F}_{p}(t) / \mathbb{F}_{p}$. In this case, $\mathbb{F}_{p}(t) \neq \mathbb{F}_{p}[t]$.
Let $F$ be a field. Then $$F(x) = \left\{ \left. \frac{a(x)}{b(x)} \right\vert a(x), b(x) \in F[x], b(x) \neq 0 \right\}$$ is a field with $$\begin{align*}+ &: \frac{a(x)}{b(x)} + \frac{c(x)}{d(x)} = \frac{a(x) d(x) + b(x) c(x)}{b(x) d(x)} \\ \times &: \frac{a(x)}{b(x)} \cdot \frac{c(x)}{d(x)} = \frac{a(x) c(x)}{b(x) d(x)}\end{align*}.$$$F(x) / F$ is simple and $x$ is a primitive element for $F(x) / F$. $F[x] \neq F(x)$.

Lecture 3. Algebraic extensions [l3]

Apr 6, 2026

Definition 3.1. algebraic over $F$ [1401]

Let $K / F$ be a field extension.

An element $\alpha \in K$ is said to be algebraic over $F$ if $\alpha$ is a root of some nonzero polynomial in $F[x]$.
An element $\alpha \in K$ is called transcendental over $F$ if it is not algebraic.
The extension $K / F$ is called algebraic if every element of $K$ is algebraic over $F$.

Example 3.2. $\mathbb{C}, \mathbb{Q}$ and $F(t)$ [1402]

$\sqrt{ 2 }, \sqrt[3]{ 2 }, i \in \mathbb{C}$ are algebraic over $\mathbb{Q}$.
$e, \pi \in \mathbb{C}$ are transcendental over $\mathbb{Q}$. (nontrivial to prove)
Let $F$ be a field. $t \in F(t)$ is transcendental over $F$.

Proposition 3.3. properties of $\alpha$ being algebraic over $F$ [1403]

Let $K / F$ be a field extension and $\alpha \in K$. The following are equivalent:

$\alpha \in K$ is algebraic over $F$,
the ring homomorphism $\varphi : F[x] \to K, \varphi \left( a(x) \right) = a(\alpha)$ is not injective,
$[F(\alpha) : F] < \infty$,
$[F[\alpha] : F] < \infty$,
$F(\alpha) = F[\alpha]$.

Proof. sketch [1403A]

(1) $\iff$ (2) is trivial.

(5) $\implies$ (1): Since $F(\alpha) = F[\alpha]$, there exists $c_{n}\alpha^{n} + \cdots + c_{1}\alpha + c_{0} \in F[\alpha]$ with at least one nonzero coefficient such that it equals $\alpha ^{-1} \in F(\alpha)$. Hence $$ \begin{align*} &\alpha ^{-1} = c_{n}\alpha^{n} + \cdots + c_{1}\alpha + c_{0} \\ \iff &1 = c_{n}\alpha^{n+1} + \cdots + c_{1}\alpha^{2} + c_{0}\alpha \\ \iff &0 = c_{n}\alpha^{n+1} + \cdots + c_{1}\alpha^{2} + c_{0}\alpha - 1 \in F[\alpha] \end{align*} $$ Therefore, $\alpha$ is algebraic over $F$.

(2) $\implies$ (3)(4)(5): Since $\varphi$ is not injective, $\ker \varphi$ is non-trivial. By the First Isomorphism Theorem, $F[x] / \ker\varphi \cong \mathrm{Im} \varphi$. Here, $\mathrm{Im}\varphi$ is a subring of $K$ containing $1$. Now $K$ is a field $\implies$ $\mathrm{Im} \varphi$ is an ID $\implies$ $F[x] / \ker \varphi$ is an ID $\implies$ $\ker\varphi$ is a PID, i.e., $\ker \varphi = \left( p(x) \right)$ for some $p(x) \in F[x]$ $\implies$ $p(x)$ is prime in $F[x]$ $\implies$ $p(x)$ is irreducible in $F[x]$. By Theorem 1206, $F[x] / \ker \varphi$ is a field and $$ F[x] / \ker\varphi = F + F\overline{x} + \cdots + F\overline{x}^{n-1}. $$ With $x \mapsto \alpha$ and $p(\alpha) = 0$ for $p(x) \in \ker\varphi$, we obtain $$ \mathrm{Im}\varphi = F + F\alpha + \cdots + F\alpha^{n-1}. $$ So $\mathrm{Im}\varphi$ is field extension of $F$ and $[\mathrm{Im}\varphi : F] < \infty$. By definition of $\varphi$, we have $F[\alpha] \subseteq \mathrm{Im}\varphi$. The above implies $\mathrm{Im}\varphi \subseteq F[\alpha]$. So $\mathrm{Im}\varphi = F[\alpha]$. Now $\mathrm{Im}\varphi$ is a field $\implies$ $F[\alpha] = F(\alpha)$. Therefore, $[F(\alpha) : F] < \infty$ and $[F[\alpha] : F] < \infty$.

(4) $\implies$ (1): $[F[\alpha] : F] < \infty$ implies that $F[\alpha] = F + F\alpha + \dots + F \alpha^{n-1}$ for some $n \in \mathbb{Z}_{\ge 0}$. Hence $1 \in F[\alpha]$ can be written as $1 = c_{0} + c_{1}\alpha + \cdots + c_{n-1}\alpha^{n-1}$, i.e., $c_{0} - 1 + c_{1}\alpha + \dots + c_{n-1}\alpha^{n-1} = 0$. So $\alpha$ is algebraic over $F$.

(3) $\implies$ (4): $F(\alpha) \subseteq F[\alpha] \subseteq F$. Then use Theorem 1207.

MATH118C Real Analysis [118c]

Mar 30, 2026 - Present

Lecture 1. Sequences and series of functions [l1]

Apr 2, 2026

Theorem 1.1. uniform convergence allows switching limits [br711]

Suppose $f_{n} \to f$ uniformly on a set $E$ in a metric space. Let $x$ be a limit point of $E$, and suppose that $$ \lim_{ t \to x } f_{n}(t) = A_{n} \quad (n = 1, 2, 3, \dots). $$ Then $\{ A_{n} \}$ converges, and $$ \lim_{ t \to x } f(t) = \lim_{ n \to \infty } A_{n}. $$ In other words, the conclusion is that $$ \lim_{ t \to x } \lim_{ n \to \infty } f_{n}(t) = \lim_{ n \to \infty } \lim_{ t \to x } f_{n}(t). $$

Proof. sketch [br711A]

Omitted.

An immediate corollary follows:

Theorem 1.2. uniform convergence of continuous fns implies continuity of the limit fn [br712]

If $\{ f_{n} \}$ is a sequence of continuous functions on $E$, and if $f_{n} \to f$ uniformly on $E$, then $f$ is continuous on $E$.

Definition 1.3. equicontinuous [br722]

A family $\mathscr{F}$ of complex functions $f$ defined on a set $E$ in a metric space $X$ is said to be equicontinuous on $E$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that $$ \lvert f(x) - f(y) \rvert < \epsilon $$ whenever $d(x, y) < \delta, x \in E, y \in E$, and $f \in \mathscr{F}$. Here $d$ denotes the metric of $X$.

It is clear that every member of an equicontinuous family is uniformly continuous.

Theorem 1.4. pointwise boundedness of $\mathscr{C}$ in a countable set [br723]

If $\{ f_{n} \}$ is a pointwise bounded sequence of complex functions on a countable set $E$, then $\{ f_{n} \}$ has a subsequence $\{ f_{n_{k}} \}$ such that $\{ f_{n_{k}}(x) \}$ converges for every $x \in E$.

Proof. sketch [br723A]

Use diagonal magic.

Let $E = \{ x_{i} \}$. Since $\{ f_{n}(x_{1}) \}$ is a bounded sequence on $\mathbb{C}$, we can find a subsequence that converges to a point of $\mathbb{C}$, denote it by $S_{1} = \{f_{1, k}(x_{1})\}$.

Now, given $\{ f_{1, k}(x_{2}) \}$, a bounded sequence on $\mathbb{C}$, we can again construct a subsequence that converges, denote it by $S_{2} = \{ f_{2, k}(x_{2}) \}$.

Similarly, we do the same for every $x_{i}$, and thus obtain sequences $S_{1}, S_{2}, S_{3}, \dots$, which we represent by the array $$ \begin{align*} &S_{1}: \ f_{1,1} \; f_{1, 2} \; f_{1, 3} \; f_{1, 4} \; \dots \\ &S_{2}: \ f_{2,1} \; f_{2, 2} \; f_{2, 3} \; f_{2, 4} \; \dots \\ &S_{3}: \ f_{3,1} \; f_{3, 2} \; f_{3, 3} \; f_{3, 4} \; \dots \\ &\dots \dots \dots \dots \dots \dots \dots \dots \end{align*} $$ Then, $\{ f_{n,n} \}$ is the desired subsequence of $\{ f_{n} \}$.

Theorem 1.5. compactness and uniform convergence implies equicontinuity [br724]

If $K$ is a compact metric space, if $f_{n} \in \mathscr{C}(K)$ for $n = 1, 2, 3, \dots$, and if $\{ f_{n} \}$ converges uniformly on $K$, then $\{ f_{n} \}$ is equicontinuous on $K$.

Proof. sketch [br724A]

~~搬砖题~~

$\{ f_{n} \}$ converges uniformly on $K$ $\implies$ for all $\epsilon > 0$, there exists an integer $N > 0$ such that for any $n, m > N$ we have $\lVert f_{n} - f_{m} \rVert < \epsilon$.

Pick any $k > N$. $f_{k} \in \mathscr{C}(K) \implies$ $f_{k}$ continuous on $K$. Now $K$ is compact $\implies$ $f_{k}$ is uniformly continuous on $K$. So there exists $\delta > 0$ such that for any $x, y \in K$ with $d(x, y) < \delta$, we have $\lvert f_{k}(x) - f_{k}(y) \rvert < \epsilon$. Then for any $n > N$, $$ \lvert f_{n}(x) - f_{n}(y) \rvert = \lvert (f_{n}(x) - f_{k}(x)) + (f_{k}(x) - f_{k}(y)) + (f_{k}(y) - f_{n}(y)) \rvert < 3\epsilon. $$ For $1 \leq n \leq N$, $f_{n}$ is uniformly continuous, just take $\delta' = \min_{1 \leq i \leq N}\delta_{i} > 0$, and our final choice of $\delta^*$ is $\min \{ \delta, \delta' \}$, which is positive.

Theorem 1.6. [br725]

If $K$ is compact, if $f_{n} \in \mathscr{C}(K)$ for $n = 1, 2, 3, \dots$, and if $\{ f_{n} \}$ is pointwise bounded and equicontinuous on $K$, then

$\{ f_{n} \}$ is uniformly bounded on $K$,
$\{ f_{n} \}$ contains a uniformly convergent subsequence.

Proof. sketch [br725A]

$\{ f_{n} \}$ is equicontinuous on $K$ $\implies$ for all $\epsilon > 0$, there exists $\delta > 0$ such that any $x, y \in K$ with $d(x, y) < \delta$ implies $\lvert f_{n}(x) - f_{n}(y) \rvert < \epsilon$ for all $n$.

For all $x \in K$, let $B_{x}$ denote the neighborhood of $x$ with radius $\delta$. Then $\{ B_{x} \}$ is an open cover of $K$. Since $K$ is compact, there exists a finite set of points $\{ p_{1}, \dots, p_{m} \}$ such that $\{ B_{p_{i}} \}$ is a finite subcover of $K$.

Now consider $B_{p_{i}}$. Let $M_{i}$ be an upper bound of $\{ \lvert f_{n}(p_{i}) \rvert \}$. Then for all $x \in B_{p_{i}}$, $|f_{n}(x)| < |f_{n}(p_{i})| + \epsilon \leq M_{i} + \epsilon$ for all $n$. Hence $M_{i} + \epsilon$ is a global upper bound of $\{ f_{n} \}$ on $B_{p_{i}}$.

Take $M = \max_{i} M_{i} + \epsilon < \infty$, then for all $x \in K$, $\lvert f_{n}(x) \rvert < M$ for all $n$. So $\{ f_{n} \}$ is uniformly bounded on $K$.
Choose any countable dense set $E$ on $K$. Then there exists a subsequence $\{ f_{n_{k}} \}$ of $\{ f_{n} \}$ such that $\{ f_{n_{k}} \}$ converges at every point of $E$. Let $\{ g_{k} \}$ denote $\{ f_{n_{k}} \}$.

Use the same $\{ B_{p_{i}} \}$ as in the first part. Pick any $x \in E$, then $x \in B_{p_{i}}$ for some $i$. Then $\{ g_{k}(x) \}$ converges, i.e., there exists a positive integer $N_{i}$ such that any $n, m > N_{i}$ means $\lvert g_{n}(x) - g_{m}(x) \rvert < \epsilon / 3$. Now for any $y \in B_{p_{i}} \cap E$, we have that for any $n, m > N_{i}$, $$\lvert g_{n}(y) - g_{m}(y) \rvert = \lvert \left( g_{n}(y) - g_{n}(x) \right) + \left( g_{n}(x) - g_{m}(x) \right) + \left( g_{m}(x) - g_{m}(y) \right) \rvert < \epsilon$$ Now choose $N = \max_{i} N_{i} < \infty$. For any $x \in E$, $n, m > N$ implies $\lvert g_{n}(x) - g_{m}(x) \rvert < \epsilon$. Hence $\{ g_{k} \}$ is uniformly convergent. Therefore, $\{ f_{n} \}$ contains a uniformly convergent subsequence.